The function $$ f(x)=\frac{4 x+7}{x-6} $$ is one-to-one. a. Find an equation for $$ f^{-1}(x) $$, the inverse function. b. Verify that your equation is correct by showing that $$ f\left(f^{-1}(x)\right)=x $$ and $$ f^{-1}(f(x))=x $$

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### Problem Overview

Given the function $$ f(x) = \frac{4x + 7}{x - 6} $$, we are to:

a. **Find the inverse function** $$ f^{-1}(x) $$.

b. **Verify the inverse** by showing that $$ f(f^{-1}(x)) = x $$ and $$ f^{-1}(f(x)) = x $$.

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### Part a: Finding the Inverse Function $$ f^{-1}(x) $$

To find the inverse function, $$ f^{-1}(x) $$, follow these steps:

1. **Start with the original function:**
   $$
   y = \frac{4x + 7}{x - 6}
   $$

2. **Swap $$ x $$ and $$ y $$:**
   $$
   x = \frac{4y + 7}{y - 6}
   $$

3. **Solve for $$ y $$:**
   $$
   x(y - 6) = 4y + 7
   $$
   $$
   xy - 6x = 4y + 7
   $$
   $$
   xy - 4y = 6x + 7
   $$
   $$
   y(x - 4) = 6x + 7
   $$
   $$
   y = \frac{6x + 7}{x - 4}
   $$

4. **Write the inverse function:**
   $$
   f^{-1}(x) = \frac{6x + 7}{x - 4}
   $$

---

### Part b: Verifying the Inverse Function

To confirm that $$ f^{-1}(x) $$ is indeed the inverse of $$ f(x) $$, we need to show that:

1. $$ f(f^{-1}(x)) = x $$
2. $$ f^{-1}(f(x)) = x $$

#### Verification 1: $$ f(f^{-1}(x)) = x $$

1. **Substitute $$ f^{-1}(x) $$ into $$ f $$:**
   $$
   f\left(f^{-1}(x)\right) = f\left(\frac{6x + 7}{x - 4}\right) = \frac{4\left(\frac{6x + 7}{x - 4}\right) + 7}{\left(\frac{6x + 7}{x - 4}\right) - 6}
   $$

2. **Simplify the numerator:**
   $$
   4\left(\frac{6x + 7}{x - 4}\right) + 7 = \frac{24x + 28}{x - 4} + 7 = \frac{24x + 28 + 7(x - 4)}{x - 4} = \frac{31x}{x - 4}
   $$

3. **Simplify the denominator:**
   $$
   \left(\frac{6x + 7}{x - 4}\right) - 6 = \frac{6x + 7 - 6(x - 4)}{x - 4} = \frac{6x + 7 - 6x + 24}{x - 4} = \frac{31}{x - 4}
   $$

4. **Combine numerator and denominator:**
   $$
   f\left(f^{-1}(x)\right) = \frac{\frac{31x}{x - 4}}{\frac{31}{x - 4}} = \frac{31x}{31} = x
   $$

**Thus, $$ f(f^{-1}(x)) = x $$.**

#### Verification 2: $$ f^{-1}(f(x)) = x $$

1. **Substitute $$ f(x) $$ into $$ f^{-1} $$:**
   $$
   f^{-1}\left(f(x)\right) = f^{-1}\left(\frac{4x + 7}{x - 6}\right) = \frac{6\left(\frac{4x + 7}{x - 6}\right) + 7}{\left(\frac{4x + 7}{x - 6}\right) - 4}
   $$

2. **Simplify the numerator:**
   $$
   6\left(\frac{4x + 7}{x - 6}\right) + 7 = \frac{24x + 42}{x - 6} + 7 = \frac{24x + 42 + 7(x - 6)}{x - 6} = \frac{31x}{x - 6}
   $$

3. **Simplify the denominator:**
   $$
   \left(\frac{4x + 7}{x - 6}\right) - 4 = \frac{4x + 7 - 4(x - 6)}{x - 6} = \frac{4x + 7 - 4x + 24}{x - 6} = \frac{31}{x - 6}
   $$

4. **Combine numerator and denominator:**
   $$
   f^{-1}\left(f(x)\right) = \frac{\frac{31x}{x - 6}}{\frac{31}{x - 6}} = \frac{31x}{31} = x
   $$

**Thus, $$ f^{-1}(f(x)) = x $$.**

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### Conclusion

Both verifications confirm that $$ f^{-1}(x) = \frac{6x + 7}{x - 4} $$ is indeed the inverse function of $$ f(x) = \frac{4x + 7}{x - 6} $$.

### Final Answers

**a.** The inverse function is
$$
f^{-1}(x) = \frac{6x + 7}{x - 4}
$$

**b.** Substituting shows that $$ f(f^{-1}(x)) = x $$ and $$ f^{-1}(f(x)) = x $$, confirming the inverse is correct.
The inverse function is $$ f^{-1}(x) = \frac{6x + 7}{x - 4} $$.

The function \( f(x)=\frac{4 x+7}{x-6} \) is one-to-one. a. Find an equation for \( f^{-1}(x) \), the inverse function. b. Verify that your equation is correct by showing that \( f\left(f^{-1}(x)\right)=x \) and \( f^{-1}(f(x))=x \)

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